Theorem freq2 4348

Description: Equality theorem for the well-founded predicate. (Contributed by NM, 3-Apr-1994.)

Assertion

Ref	Expression
freq2	⊢ (𝐴 = 𝐵 → (𝑅 Fr 𝐴 ↔ 𝑅 Fr 𝐵))

Proof of Theorem freq2

Dummy variable 𝑠 is distinct from all other variables.

Step	Hyp	Ref	Expression
1		frforeq2 4347	. . 3 ⊢ (𝐴 = 𝐵 → ( FrFor 𝑅𝐴𝑠 ↔ FrFor 𝑅𝐵𝑠))
2	1	albidv 1824	. 2 ⊢ (𝐴 = 𝐵 → (∀𝑠 FrFor 𝑅𝐴𝑠 ↔ ∀𝑠 FrFor 𝑅𝐵𝑠))
3		df-frind 4334	. 2 ⊢ (𝑅 Fr 𝐴 ↔ ∀𝑠 FrFor 𝑅𝐴𝑠)
4		df-frind 4334	. 2 ⊢ (𝑅 Fr 𝐵 ↔ ∀𝑠 FrFor 𝑅𝐵𝑠)
5	2, 3, 4	3bitr4g 223	1 ⊢ (𝐴 = 𝐵 → (𝑅 Fr 𝐴 ↔ 𝑅 Fr 𝐵))

Syntax hints:   → wi 4   ↔ wb 105  ∀wal 1351   = wceq 1353   FrFor wfrfor 4329   Fr wfr 4330

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-io 709  ax-5 1447  ax-7 1448  ax-gen 1449  ax-ie1 1493  ax-ie2 1494  ax-8 1504  ax-10 1505  ax-11 1506  ax-i12 1507  ax-bndl 1509  ax-4 1510  ax-17 1526  ax-i9 1530  ax-ial 1534  ax-i5r 1535  ax-ext 2159

This theorem depends on definitions:  df-bi 117  df-tru 1356  df-nf 1461  df-sb 1763  df-clab 2164  df-cleq 2170  df-clel 2173  df-nfc 2308  df-ral 2460  df-in 3137  df-ss 3144  df-frfor 4333  df-frind 4334

This theorem is referenced by:  weeq2  4359